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Stels [109]
3 years ago
15

Convert the decimal number 61078 to binary by using sum-of-weights method​

Physics
1 answer:
marshall27 [118]3 years ago
5 0

Answer:

1110111010010110

Explanation:

I am not able to upload the working out using the sum of weights method sorry

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The Force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What i
shtirl [24]

Answer: The coefficient of kinetic friction is μ = 0.6

Explanation:

For an object of mass M, the weight is:

W = M*g

where g is the gravitational acceleration: g = 9.8m/s^2

And the friction force between this object and the surface can be written as:

F = W*μ

where μ is the coefficient of friction (kinetic if the object is moving, and static if the object is not moving, usually the static coefficient is larger)

In this case, the weight is:

W = 20N

And the friction force is:

F = 12N

Replacing these values in the equation for the friction force we get:

12N = 20N*μ

(12N/20N) = μ = 0.6

The coefficient of kinetic friction is μ = 0.6

7 0
3 years ago
Read 2 more answers
Net force of 200 newtons is applied to a wagon for 3 seconds
AveGali [126]

Momentum of the wagon increases by (200 x 3)

                                                     = 600 newton-sec

                                                     = 600 kg-m/sec
 
6 0
4 years ago
A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentu
tekilochka [14]

Answer: 0.43\ rad/s

Explanation:

Given

Mass of child m=34\ kg

speed of child is v=2.8\ m/s

Moment of inertia of merry go round is I=510\ kg.m^2

radius r=2.31\ m

Conserving the angular momentum

\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s

7 0
3 years ago
Two identical vertical springs S1 and S2 have masses m1 = 400 g and m2 = 800 g attached to them. If m1 causes spring S1 to stret
OlgaM077 [116]

Answer:

potential energy = mgh

= 400÷1000 × 10× 4÷100

= 0.4 × 10 × 0.04

=4/10 ×10×4/100

= 4/10 × 4/10

=16/100

= 0.16 joules

m1 (400) stretches 4cm

m1 (100g) stretches 1cm

so, m2(800g) stretches 8 cm

potential energy of m2 = mgh

= 800/1000 ×10×8/100

= 0.8 × 0.8

=8/10 ×8/10

= 64/100

=0.64 joules

Ratio of s1 to s2

16/100 ÷ 64/100

= 1:4 ( answer)

6 0
3 years ago
A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20
alexandr402 [8]

Answer:

the amplitude of  the subsequent oscillations is 0.11  m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, m_{1} = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, m_{2} = 200 g = 0.2 kg

the speed of glider,  v_{2} = 170 cm/s = 1.7 m/s

the amplitude of  the subsequent oscillations is A = 0.11  m

according to mechanical enery equation, we have

A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}

where

A is the amplitude and  v_{f} is the final speed.

to find v_{f}, we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

P_{f} = P_{i}

(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}

v_{1} = 0, thus

(0.75+0.2)v_{f} = (0.75)(0)+(0.2)(1.7)

0.95 v_{f} = 0.34

v_{f} = 0.36 m/s

Now we can calculate the amplitude

A = \sqrt{\frac{0.75 +0.2 }{10} }0.36

A = 0.11  m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π\sqrt{\frac{m_{1}+m_{2}  }{k} }

T = 2π\sqrt{\frac{0.75+0.2  }{10} }

T = 1.94 s

7 0
3 years ago
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